Question

In a town of 10,000 families it was found that 40% families buy newspaper A, 20% buy newspaper B and 10% buy newspaper C, 5% buy A and B, 3% buy B and C and 4% buy A and C. If 2% buy all the three newspapers, then number of families which buy A only is ________.

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Answer 1

n(A) = 40% of 10,000 = 4,000

n(B) = 20% of 10,000 = 2,000

n(C) = 10% of 10,000 = 1,000

n (A$\cap$B) = 5% of 10,000 = 500

n (B$\cap$C) = 3% of 10,000 = 300

n(C$\cap$A) = 4% of 10,000 = 400

n(A$\cap$B$\cap$C)= 2% of 10,000 = 200

Representation of data on Venn diagram

We want to find n(A$\cap$B^c$\cap$C^c) = n[A∩ ${(B\cup C)}^c$]

= n(A) - n[A$\cap$ (B$\cup$C)] = n(A) - n[(A$\cap$B) $\cup$ (A$\cap$C)]

= n(A) - [n(A$\cap$B) + n(A$\cap$C) - n(A$\cap$B$\cap$C)]

= 4000 - [500 + 400 - 200] = 4000 - 700 = 3300.

OR

From Venn diagram, we observe that

Number of families who buy A only = Number of families who buy A - Number of families who buy A and B - Number of families who buy A and C + Number of families who buy A, B and C

{ since, we subtracted number of families who buy A, B and C twice so, add it one time }

= 4000 - 500 - 400 + 200 = 3300